Weeks 6
8088/8086 Microprocessor
Programming
Shift
Target register or memory
C
0
SHL
equivalent
C
SAL
0
C
SHR
0
C
SAR
Sign Bit
2
Examples
Examples SHL AX,1
SAL DATA1, CL ; shift count is a modulo-32 count
Ex.
Ex.
Ex.
; Multiply AX by 10
SHL AX, 1
MOV BX, AX
MOV CL,2
SHL AX,CL
ADD AX, BX
What are the results of SAR CL, 1 if CL initially contains B6H?
What are the results of SHL AL, CL if AL contains 75H
and CL contains 3?
3
Rotate
C
Target register or memory
RCL
C
ROL
C
RCR
C
ROR
Ex.
What is the result of ROL byte ptr [SI], 1 if this memory location 3C020
contains 41H?
What is the result of ROL word ptr [SI], 8 if this memory location 3C020
contains 4125H?
4
Example
Write a program that counts the number of 1’s in a
byte and writes it into BL
DATA1 DB 97
SUB
MOV
MOV
AGAIN: ROL
JNC
INC
NEXT: DEC
JNZ
NOP
; 61h
BL,BL
;clear BL to keep the number of 1s
DL,8
;rotate total of 8 times
AL,DATA1
AL,1
;rotate it once
NEXT
;check for 1
BL
;if CF=1 then add one to count
DL
;go through this 8 times
AGAIN
;if not finished go back
5
BCD and ASCII Numbers
•
BCD (Binary Coded Decimal)
– Unpacked BCD: One byte per digit
– Packed BCD: 4 bits per digit (more efficient in storing data)
•
ASCII to unpacked BCD conversion
– Keyboards, printers, and monitors all use ASCII.
– Digits 0 to 9 are represented by ASCII codes 30 – 39.
•
Example. Write an 8086 program that displays the packed BCD
number in register AL on the system video monitor
– The first number to be displayed should be the MS Nibble
– It is found by masking the LS Nibble and then rotating the MS Nibble
into the LSD position
– The result is then converted to ASCII by adding 30h
– The BIOS video service is then called to display this result.
6
ASCII Numbers Example
MOV BL,AL; save
AND AL,F0H
MOV CL,4
ROR AL,CL
ADD AL,30H
MOV AH,0EH
INT 10H ;display single character
MOV AL,BL; use again
AND AL,0FH
ADD AL,30H
INT 10H
INT 20H
; RETURN TO DOS
C
7
Example
• Write an 8086 program that adds two packed BCD
numbers input from the keyboard and computes and
displays the result on the system video monitor
• Data should be in the form 64+89= The answer 153
should appear in the next line.
#
0
?
1
6
2
4
3
+
4
8
5
9
6
=
7
8
Example Continued
Mov dx, offset bufferaddress
Mov ah,0a
Mov si,dx
Mov byte ptr [si], 6
Int 21
Mov ah,0eh
Mov al,0ah
Int 10
; BIOS service 0e line feed position cursor
sub byte ptr[si+2], 30h
sub byte ptr[si+3], 30h
sub byte ptr[si+5], 30h
sub byte ptr[si+6], 30h
6
?
6
0
1
2
4
3
+
4
8
5
9
6
=
7
Mov cl,4
Rol byte ptr [si+3],cl
Rol byte ptr [si+6],cl
Ror word ptr [si+5], cl
Ror word ptr [si+2], cl
Mov al, [si+3]
Add al, [si+6]
Daa
Mov bh,al
Jnc display
Mov al,1
Call display
Mov al,bh
Call display
Int 20
9
Flag Control Instructions
SF
•
•
ZF
AF
PF
LAHF Load AH from flags (AH) ← (Flags)
SAHF Store AH into flags (Flags) ← (AH)
CF
Bulk manipulation
of the flags
– Flags affected: SF, ZF, AF, PF, CF
•
•
•
•
•
CLC Clear Carry Flag (CF) ← 0
STC Set Carry Flag (CF) ← 1
CLI Clear Interrupt Flag (IF) ← 0
STI Set interrupt flag (IF) ← 1
Example (try with debug)
Individual
manipulation of
the flags
LAHF
MOV AX,0000
ADD AX,00
SAHF
– Check the flag changes!
10
Compare
Unsigned Comparison
Signed Comparison
Comp
Operands
CF
ZF
Comp
Operands
ZF
SF,OF
Dest >
source
0
0
Dest >
source
0
SF=OF
Dest =
source
0
1
Dest =
source
1
x
Dest <
source
1
0
Dest <
source
0
SF<>OF
11
Compare Example
DATA1
…
DW
235Fh
MOV AX, CCCCH
CMP AX, DATA1
JNC OVER
SUB AX,AX
OVER: INC DATA1
CCCC – 235F = A96D => Z=0, CF=0 =>
CCCC > DATA1
12
Compare (CMP)
For ex: CMP CL,BL ; CL-BL; no modification on neither operands
Write a program to find the highest among 5 grades and write it in DL
DATA
DB
51, 44, 99, 88, 80
MOV CX,5
MOV BX, OFFSET DATA
SUB AL,AL
AGAIN: CMP AL,[BX]
JA NEXT
MOV AL,[BX]
NEXT: INC BX
LOOP AGAIN
MOV DL, AL
;13h,2ch,63h,58h,50h
;set up loop counter
;BX points to GRADE data
;AL holds highest grade found so far
;compare next grade to highest
;jump if AL still highest
;else AL holds new highest
;point to next grade
;continue search
13
Jump Instructions
•
Unconditional vs
conditional jump
14
Conditional Jump
These flags are based on general comparison
Mnemonic
Description
Flags/Registers
JZ
Jump if ZERO
ZF = 1
JE
Jump if EQUAL
ZF = 1
JNZ
Jump if NOT ZERO
ZF = 0
JNE
Jump if NOT EQUAL
ZF = 0
JC
Jump if CARRY
CF = 1
JNC
Jump if NO CARRY
CF = 0
JCXZ
Jump if CX = 0
CX = 0
JECXZ
Jump if ECX = 0
ECX = 0
15
Conditonal Jump based on flags
Mnemonic
Description
Flags/Registers
JS
JUMP IF SIGN (NEGATIVE)
SF = 1
JNS
JUMP IF NOT SIGN (POSITIVE)
SF = 0
JP
Jump if PARITY EVEN
PF = 1
JNP
Jump if PARITY ODD
PF = 0
JO
JUMP IF OVERFLOW
OF = 1
JNO
JUMP IF NO OVERFLOW
OF = 0
16
Jump Based on Unsigned Comparison
These flags are based on unsigned comparison
Mnemonic
Description
Flags/Registers
JA
Jump if above op1>op2
CF = 0 and ZF = 0
JNBE
Jump if not below or equal
op1 not <= op2
CF = 0 and ZF = 0
JAE
Jump if above or equal
op1>=op2
CF = 0
JNB
Jump if not below
op1 not <opp2
CF = 0
JB
Jump if below op1<op2
CF = 1
JNAE
Jump if not above nor equal
op1< op2
CF = 1
JBE
Jump if below or equal
op1 <= op2
CF = 1 or ZF = 1
JNA
Jump if not above
op1 <= op2
CF = 1 or ZF = 1
17
Jump Based on Signed Comparison
These flags are based on signed comparison
Mnemonic
Description
Flags/Registers
JG
Jump if GREATER op1>op2
SF = OF AND ZF = 0
JNLE
Jump if not LESS THAN or equal op1>op2
SF = OF AND ZF = 0
JGE
Jump if GREATER THAN or equal op1>=op2
SF = OF
JNL
Jump if not LESS THAN op1>=op2
SF = OF
JL
Jump if LESS THAN op1<op2
SF <> OF
JNGE
Jump if not GREATER THAN nor equal
op1<op2
SF <> OF
JLE
Jump if LESS THAN or equal op1 <= op2
ZF = 1 OR SF <> OF
JNG
Jump if NOT GREATER THAN op1 <= op2
ZF = 1 OR SF <> OF
18
Control Transfer Instructions (conditional)
• It is often necessary to transfer the program
execution.
– Short
• A special form of the direct jump: “short jump”
• All conditional jumps are short jumps
• Used whenever target address is in range +127 or –128 (single
byte)
• Instead of specifying the address a relative offset is used.
19
Short Jumps
•Conditional Jump is a two byte instruction.
•In a jump backward the second byte is the 2’s complement of the
displacement value.
•To calculate the target the second byte is added to the IP of the instruction
after the jump.
Ex:
000D ADD AL,[BX]
000F INC BX
0010 DEC CX
Short Jump 0013 + FA (-6)
= 0D
0011 JNZ FA
0013
20
Hello2.exe
SJ Example
.model small
.stack 100h
.data
org 0010
message1 db "You now have a small letter
entered !",0dh,0ah,'$'
org 50
message2 db "You have NON small letters
",0dh,0ah,'$'
.code
main proc
mov ax,@data
mov ds,ax
mov ah,00h
int 16h
cmp al,61h
jb next
Cmp al,7Ah
ja next
mov ah,09h
mov dx,offset message1
mov ah,09h
int 21h
int 20h
next: mov dx,offset message2
mov ah,09h
int 21h
mov ax,4C00h
int 21h
main endp
end main
21
A Simple Example Program finds the sum
•
Write a program that adds 5 bytes of data and saves the result. The
data should be the following numbers: 25,12,15,10,11
.model small
Again: add al,[bx]
.stack 100h
inc bx
.data
dec cx
Data_in DB 25,12,15,10,11
jnz Again
Sum DB ?
mov sum,al
.code
mov ah,4Ch
main proc far
INT 21H
mov ax, @Data
Main
endp
mov ds,ax
end main
mov cx,05h
mov bx,offset data_in
mov al,0
22
Example Output
23
Unconditional Jump
™Short Jump: jmp short L1 (8 bit)
™Near Jump: jmp near ptr Label
If the control is transferred to a memory location within the current
code segment (intrasegment), it is NEAR. IP is updated and CS
remains the same
¾The displacement (16 bit) is added to the IP of the instruction following jump
instruction. The displacement can be in the range of –32,768 to 32,768.
¾The target address can be register indirect, or assigned by the label.
¾Register indirect JMP: the target address is the contents of two memory
locations pointed at by the register.
¾Ex: JMP [SI] will replace the IP with the contents of the memory locations
pointed by DS:DI and DS:DI+1 or JMP [BP + SI + 1000] in SS
™Far Jump: If the control is transferred to a memory location outside the
current segment. Control is passing outside the current segment both CS and IP
have to be updated to the new values. ex: JMP FAR PTR label = EA 00 10 00 20
jmp far ptr Label ; this is a jump out of the current segment.
24
Near Jump
Jumps to the specified IP with +/- 32K distance from
the next instruction following the jmp instruction
25
Far Jump
Jumps to the specified CS:IP
26
XLAT
•
•
Adds the contents of AL to BX and uses the resulting offset to point
to an entry in an 8 bit translate table.
This table contains values that are substituted for the original value
in AL.
The byte in the table entry pointed to by BX+AL is moved to AL.
•
XLAT [tablename] ; optional because table is assumed at BX
•
Table db ‘0123456789ABCDEF’
•
Mov AL,0A; index value
Mov bx,offset table
Xlat; AL=41h, or ‘A’
27
Subroutines and Subroutine Handling Functions
9A subroutine is a special
segment of a program that can
be called for execution from
any point in the program
9A RET instruction must be
included at the end of the
subroutine to initiate the return
sequence to the main program
environment
Examples. Call 1234h
Call BX
Call [BX]
Two calls
•intrasegment
•intersegment
28
Calling a NEAR proc
9The CALL instruction and the subroutine it calls are in
the same segment.
9Save the current value of the IP on the stack.
9load the subroutine’s offset into IP (nextinst + offset)
Calling Program
Subroutine
Main proc
001A: call sub1
001D: inc ax
.
Main endp
sub1 proc
0080: mov ax,1
…
ret
sub1 endp
Stack
1ffd
1D
1ffe
00
1fff
(not used)
29
Calling a FAR proc
9The CALL instruction and the subroutine it calls are in
the “Different” segments.
9Save the current value of the CS and IP on the stack.
9Then load the subroutine’s CS and offset into IP.
Calling Program
Main proc
1FCB:001A: call far ptr sub1
1FCB:001F: inc ax
…
…
Main endp
Subroutine
sub1 proc far
4EFA:0080: mov ax,1
….
….
ret (retf opcode generated)
sub1 endp
Opcode 8000 FA4E
Stack
1ffb
1F
1ffc
00
1ffd
CB
1ffe
1F
1fff
N/A
I
P
S
E
G
30
Example on Far/Near Procedure Calls
0350:1C00 Call FarProc
0350:1C05 Call NearProc
0350:1C08 nop
1ff0
08
1ffa
1C
1ffb
05
1ffc
1C
1ffd
50
1ffe
03
1fff
X
31
Nested Procedure Calls
A subroutine may itself call other subroutines.
Example:
000A
000C
…
0030
0040
subr2 proc
nop
…
call subr3
ret …
1ff0
60
main endp
subr2 endp
1ffa
00
subr1 proc
nop
…
call subr2
ret …
subr3 proc
nop
…
nop
ret
subr3 endp
1ffb
40
1ffc
00
1ffd
0c
1ffe
00
1fff
X
main proc
call subr1
mov ax,…
0050
0060
subr1 endp
0070
0079
007A
Q: show the
stack contents
at 0079?
Do NOT overlap Procedure Declarations
32
Push and Pop Instructions
Push S (16/32 bit or Mem)
(SP) ← (SP) - 2
((SP)) ← (S)
Pop D (16/32 bit or Mem)
(D) ← ((SP))
(SP) ← (SP) + 2
33
Loop and Loop Handling Instructions
34
Loop
35
Nested Loops
single Loop
MOV CX,A
BACK: …
…
…
…
LOOP BACK
How many
times will the
loop execute,
if JCXZ wasn’t
there
Nested Loops
MOV CX,A
OUTER: PUSH
NOP CX
MOV CX, 99
INNER: NOP
…
…
…
LOOP INNER
NOP CX
POP
LOOP OUTER
MOV CX,0
DLOOP: JCXZ SKIP ;guarding
BACK: MUL AX,2H
ADD AX,05H
LOOP BACK
SKIP: INC AX; if CX=0
36
INT
INT operates similar to Call
™Processor first pushes the flags
™Trace Flag and Interrupt-enable flags are cleared
™Next the processor pushes the current CS register onto the stack
™Next the IP register is pushed
Example: What is the sequence of events for INT 08? If it generates a CS:IP
of 0100:0200. The flag is 0081H.
SP-6
00
MEMORY / ISR table
SP-5
02
00020
10
SP-4
00
00021
00
SP-3
01
00022
80
SP-2
81
00023
05
SP-1
00
SP initial
I
P
S
E
G
0580:
0010
37
IRET
•IRET must be used for special handling of the stack.
•Must be used at the end of an ISR
SP-6
00
SP-5
02
SP-4
00
SP-3
01
SP-2
81
SP-1
00
Return address +
flags are loaded
SP initial
38
String Instructions
80x86 is equipped with special instructions to handle string
operations
String: A series of data words (or bytes) that reside in
consecutive memory locations
Operations: move, scan, compare
String Instruction:
Byte transfer, SI or DI increment or decrement by 1
Word transfer, SI or DI increment or decrement by 2
DWord transfer, SI or DI increment or decrement by 4
39
String Instructions - D Flag
The Direction Flag: Selects the auto increment D=0 or
the auto decrement D=1 operation for the DI and SI registers
during string operations. D is used only with strings
CLD Æ Clears the D flag / STD Æ Sets the D flag
40
String Instructions
41
Repeat String REP
Basic string operations must be repeated in order
to process arrays of data; this is done by inserting a
repeat prefix.
42
Example. Find and replace
•
Write a program that scans the name “Mr.Gohns” and replaces the
“G” with the letter “J”.
Data1 db 'Mr.Gones','$‘
.code
mov es,ds
cld ;set auto increment bit D=0
mov di, offset data1
mov cx,09; number of chars to be scanned
mov al,'G'; char to be compared against
repne SCASB; start scan AL =? ES[DI]
jne Over; if Z=0
dec di; Z=1
mov byte ptr[di], 'J'
Over: mov ah,09
mov dx,offset data1
int 21h; display the resulting String
Search.exe
search.asm
43
Strings into Video Buffer
Fill the Video Screen with a value
CLD
MOV
MOV
MOV
MOV
MOV
REP
Clear.exe
AX,0B800H
ES,AX
DI,0
CX,2000H
AL,20h
STOSW
44
Example. Display the ROM BIOS Date
•
•
•
•
•
Write an 8086 program that searches the BIOS ROM for its
creation date and displays that date on the monitor.
If a date cannot be found display the message “date not found”
Typically the BIOS ROM date is stored in the form xx/xx/xx
beginning at system address F000:FFF5
Each character is in ASCII form and the entire string is terminated
with the null character (00)
Add a ‘$’ character to the end of the string and make it ready for
DOS function 09, INT 21
Date.asm
45